# Calc, integration, and me forgetting it

Registered User regular
edited September 2007
I'm in Calc 2, and it's been over a year and a half since I had any math class. Thus, I've forgotten even the basics.

What we're working on now is integration by parts. Example problem 1:

integral of (216(x^2)cos(6x))

So what I did so far was set
u=216(x^2)        dv=cos(6x)
du=432x            v=(sin(6x)/6)

I'm pretty sure it's correct so far, but I honestly can't remember how to do it. Anyways, I now have

216(x^2)(sin(6x)/6)- int[(sin(6x)/6)(432x)dx]

Which, when solved, is equal to

36(x^2)(sin(6x))+6(x^2)(cos(6x))

which is wrong. Where did I mess up?

Food? on

## Posts

• Registered User regular
edited September 2007
it's been a while for me too, but offhand I'd say switch what you have for u and dv, because it's much easier to find the derivative of cos6x than the antiderivative...also when you find the antiderivative of 216x^2, it reduces to 72x^3 which is a pretty easy number to work with.

EDIT: given that, let me know if you can work out the rest, if not I'll try to rack my brains for the rest. Looking at it though I think you'll probably have to do IBP once more to get the answer.

EDIT2: oops maybe I was wrong, this should not be that difficult to solve. Hold on a second.

EDIT3: OK I'm pretty much stumped here. Too many cobwebs in my brain...Reasonably certain that it's the second part of your solution that is incorrect though. Sorry I couldn't help more.

Chief1138 on
• Registered User regular
edited September 2007
Are you sure you copied this problem right? It seems weird to have that random 216 in front of everything. Since that's a number, you can just take it out of the integral, since integral[ a f(x) dx ] = a integral[ f(x) dx] (since integrals are just sums).

In problems like this you want to use u=x^p and dv=trig(x). That way you eliminate powers of x to get something you know how to integrate.

So here you have u=x^2, dv=cos(6x) dx
So, du =2x dx, v = 1/6 sin(6x)

So u*v - Integral [ v du ] = x^2 sin(6x)/6 - Integral[ 2x 1/6 sin(6x) ]

Then, do integration by parts again to get the integral of cosine, which is something that's easy to do.

cfgauss on
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• Registered User regular
edited September 2007
cfgauss wrote: »
Are you sure you copied this problem right? It seems weird to have that random 216 in front of everything. Since that's a number, you can just take it out of the integral, since integral[ a f(x) dx ] = a integral[ f(x) dx] (since integrals are just sums).

In problems like this you want to use u=x^p and dv=trig(x). That way you eliminate powers of x to get something you know how to integrate.

So here you have u=x^2, dv=cos(6x) dx
So, du =2x dx, v = 1/6 sin(6x)

So u*v - Integral [ v du ] = x^2 sin(6x)/6 - Integral[ 2x 1/6 sin(6x) ]

Then, do integration by parts again to get the integral of cosine, which is something that's easy to do.

That's what I did, and WebWork still says it's incorrect. Here's what's been entered as my answer:

216((x^2)[ sin(6x)/6]+[ x/18]cos(6x))+C

The parentheses are somewhat confusing because WebWork automatically takes what you've entered and converts it into something it understands before it checks it. Why is that answer wrong?

Food? on
• Registered User regular
edited September 2007
Okay, so I had someone from my class show the answer to me, and apparently it's

36(x^2)sin(6x) +(12x)cos(6x) -(2)sin(6x) +C

but I can't figure out how. I'd really like to learn where I messed up, so I don't do it again.

Food? on
• Registered User regular
edited September 2007
I'm leaving out the factor of 216 for simplicity's sake.
int(x^2*cos(6*x),x) = x^2*sin(6*x)/6-1/6*int(2*x*sin(6*x),x)
=x^2*sin(6*x)/6-1/3*(-x*cos(6*x)/6+int(cos(6*x)/6,x))
=x^2*sin(6*x)/6-1/3*(-x*cos(6*x)/6+sin(6*x)/36)
=x^2*sin(6*x)/6+1/18*x*cos(6*x)-1/108*sin(6*x)
Multiply through by the 216 and you get the answer.

inlemur on
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• Registered User regular
edited September 2007
Just throwing this out there: Personally, I can never remember the formula for integration by parts. I always re-derive it when I need it. It comes straight from the product rule,

(uv)' = uv' + vu'.

Just integrate to get

uv = int(uv') + int(vu')

and rearrange to get

int(uv') = uv - int(vu').

Anyway,
Food? wrote: »
Anyways, I now have

216(x^2)(sin(6x)/6) - int[(sin(6x)/6)(432x)dx]

Which, when solved, is equal to

36(x^2)(sin(6x)) + 6(x^2)(cos(6x))

which is wrong. Where did I mess up?

The bolded parts are not equal. The bolded integral is another integral you need to do...by parts.

Marty81 on